Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <unordered_set>
#include <queue>

int len[100001] = { 0 };
int len2[100001] = { 0 };
int count[100001] = { 0 };
int count2[100001] = { 0 };
std::vector<int> e[100001];

int main() {
    int n, m, s, t;
    std::cin >> n >> m >> s >> t;
    s; t;

    for (int i = 0; i < m; i++) {
        int x, y;
        std::cin >> x >> y;
        e[x].push_back(y);
        e[y].push_back(x);
    }

    std::queue<std::pair<int, int>> q; // first:今の場所 second:cost
    q.push(std::make_pair(s, 0));

    int minLen = 0;
    bool isGoal = false;
    while (!q.empty() && !isGoal) {
        auto node = q.front(); q.pop();
        node.second++;
        for (auto to : e[node.first]) {
            if (len[to] && len[to] <= node.second || to == s) continue;
            auto next = node;
            len[to] = next.second;
            next.first = to;
            q.push(next);

            if (to == t) {
                minLen = next.second;
                isGoal = true;
                break;
            }
        }
    }

    // 逆方向
    q = std::queue<std::pair<int, int>>();
    q.push(std::make_pair(t, 0));
    isGoal = false;
    while (!q.empty() && !isGoal) {
        auto node = q.front(); q.pop();
        node.second++;
        // 打ち切り
        if (node.second > minLen - 2) {
            break;
        }
        for (auto to : e[node.first]) {
            if ((len2[to] && len2[to] <= node.second) || to == s || to == t) continue;

            auto next = node;
            len2[to] = next.second;
            next.first = to;
            q.push(next);

            if (to == t) {
                isGoal = true;
                break;
            }
        }
    }

    // 距離カウント
    for (auto x : len) {
        count[x]++;
    }
    for (auto x : len2) {
        count2[x]++;
    }

    // 数カウント
    long long num = 0;
    for (int i = 0; i < minLen - 1; i++) {
        long long from = count[i];
        long long to = count2[minLen - (i + 2)];
        if (!i) from = 1;
        if (minLen - (i + 2) == 0) to = 1;

        num += from * to;
    }

    std::cout << num << std::endl;
    return 0;
}